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Mastering Polynomial Factoring Techniques

Learn to factor higher degree polynomials efficiently using examples and step-by-step instructions. Practice with various exercises to enhance your skills. Use factoring by grouping and synthetic division, and discover integral zeros through graphing. Strengthen your understanding of factoring methods with detailed explanations.

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Mastering Polynomial Factoring Techniques

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  1. UNIT 1ALESSON 2 FACTORING POLYNOMIAL EXPRESSIONS

  2. Factor Review x2 + 3x – 4 x2 – 25 (x – 5)(x + 5) (x + 4)(x – 1) x2 – 9x + 8 4x2 – 20x + 25 (x – 8)(x – 1) (2x – 5)2 2x2 – x – 3 3x2 + x – 4 3x2 – 3x + 4x – 4 3x(x – 1) + 4(x – 1) (3x + 4)(x – 1) 2x2 + 2x – 3x – 3 2x(x + 1) – 3(x + 1) (2x – 3)(x + 1)

  3. Factoring Higher Degree Polynomials Do you remember how to factor polynomials such as F(x) = x3 + 6x – 4x + 1 G(x) = x4 – 7x2 + 5x – 2 or if they even factor?

  4. Factor by Grouping Sometimes, a polynomial expression can be factored by grouping. (Similar to decomposition) • Example 1 2x3+ 3x2 – 8x – 12 x2(2x + 3) – 4 (2x + 3) (x2 – 4)(2x + 3) (x – 2)(x + 2)(2x + 3)

  5. Practise 1. f(x) = x3 + x2– 9x – 9 x3 + x2 – 9x – 9 x2(x + 1) – 9 (x + 1) (x2 – 9)(x + 1) (x – 3)(x + 3)(x + 1) 2. g(x) = 4x3 + 8x2 – 25 x – 50 4x3 + 8x2 – 25x – 50 4x2(x + 2) – 25 (x + 2) (4x2 – 25)(x + 2) (2x – 5)(2x + 5)(x + 2)

  6. Complete Question 1

  7. Factoring Polynomial Expressions EXAMPLE 2: Consider the polynomial P(x) = x3 – 7x– 6 Step 1: List the integers, which divide into the constant term–6 These are called potential zeros + 1, +2, +3, +6 If P(a) = 0 then (x – a) is a factor Step 2: Test the potential zeros until you find zero then state the factor P(1) = (1)3 – 7 (1) – 6 = –12 therefore ( x – 1) is NOT a factor P(–1) = (–1)3– 7 (–1) – 6 = 0 therefore ( x+ 1) is IS a factor

  8. Coefficients of P(x) • 1 0 -7 -6 • 1 -1 -6 • 1 -1 -6 0 Synthetic Division P(x) = x3 – 7x – 6 has the factor x + 1 Step 3: Divide the polynomial with the factor you found Method 1: Using subtraction Quotient x2 – x – 6 Remainder 0 Step 4: Complete the factoring of the polynomial P(x) = x3 – 7x – 6 P(x) = (x + 1)(x2 – x – 6) P(x) = (x + 1)(x – 3)(x + 2)

  9. (-2, 0) (3, 0) (-1, 0) Check by Graphing Graph the polynomial P(x) = x3 – 7x – 6 The x-intercepts give us the zeros of the function which we can use to determine the factors (x + 2)(x + 1)(x – 3)

  10. Complete questions 2-3 on pages 3-4

  11. EXAMPLE 3: P(x) = x3 – x2 – 5x + 6 Find the zeros of P(x) by graphing and determining the x-intercepts 2 –2.3 1.3 Integral Zero ApproximateZeros

  12. -2 1 -1 -5 6 -2 -2 6 1 1 -3 0 EXAMPLE 3: P(x) =x3 – x2 – 5x + 6 From the previous graph the Integral zero is x = 2 OR P(2) = (2)3 – (2)2 – 5(2) + 6 =0so (x – 2) is a factor P(x) = (x – 2)(x2 + x – 3) Exact roots x = 2, x = – 2.3 or x = 1.3 Approximate roots

  13. Complete question 4 page 6

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